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I was messing around and I noticed that $(2^2\cdot3^3)^2 = 2^2\cdot 3^3\cdot 4^4$.

May sound strange but $108$ is a mystical number of ancient India, and was trying to deduce why, when I noticed it is $2^2\cdot 3^3$. So then I wondered what would happen if I multiplied that to $4$ to the $4^{\textrm{th}}$ power and discovered that it is $108^2$.

This got me thinking that maybe this is a universal formula that I just was not aware of... does $(a^2\cdot b^3)^2 = a^2\cdot b^3\cdot c^4$?

Pardon my ignorance, it has been many years since I've been in a math class.

Thanks.

UPDATE : my math was wrong, please disregard this question .. should of double-checked the calculation before posting

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  • $\begingroup$ I'm voting to close this question as off-topic because it is based on a false premise. $\endgroup$
    – quid
    Feb 13, 2015 at 14:49
  • $\begingroup$ What did I just read? How are the two even the same? Have you tried it on a calculator? $\endgroup$
    – Rohinb97
    Feb 13, 2015 at 14:51
  • $\begingroup$ maybe i made a mistake. sorry i will delete this. won't let me delete it, but I tried. $\endgroup$ Feb 13, 2015 at 14:52
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    $\begingroup$ Question based on a false premise. $\endgroup$
    – user65203
    Feb 13, 2015 at 14:56

3 Answers 3

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$2^2*3^3*4^4=4*27*256=27648\\(2^2*3^3)^2=108^2=11664$

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Using the basic laws of exponents, one has

$$(a^2\cdot b^3)^2=(a^2\cdot b^3) \cdot(a^2\cdot b^3) = a^2 \cdot b^3\cdot (a^2\cdot b^3)$$

and, generally speaking, the expression $(a^2\cdot b^3)$ is not typically a fourth power of another integer $c$. Of course, it is the fourth power of $\sqrt[4]{a^2\cdot b^3}$, but that's not usually an integer.

So -- no.

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$$(2^2\times 3^3)^2 = 2^23^34^4$$ $$\implies2^23^3 = 4^4 \text{, but }$$ this is impossible since LHS is odd, RHS is even

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